Counter Example about Continuous Functions

Question

(James Munkres page 104 Theorem 18.1) Let $X$ and $Y$ be topological spaces; let $f: X \rightarrow Y$. If $f$ is continuous, then for every subset $A$ of $X$, one has $f(\overline{A})\subset \overline{f(A)}$. I can follow the argument in the book. My question is why cannot one has equality here, that is, $f(\overline{A}) = \overline{f(A)}$, please? Is there an easy counter example, please? Thank you!

Answer 1

There is a nice symmetry here: $f$ is continuous iff for all $A \subset X$ we have $f[\overline{A}] \subset \overline{f[A]}$ (which means intuitively that a point "close to" $A$ has an image "close to" $f[A]$), and $f$ is closed (maps closed sets to closed sets) iff for all $A \subset X$ we have $\overline{f[A]} \subset f[\overline{A}]$.

Proof of the latter: suppose $f$ is closed, and $A \subset X$. Then $\overline{A}$ is closed in $X$, so $f[\overline{A}]$ is closed in $Y$, and it contains $f[A]$, so it also contains its closure. For the reverse, suppose we have the inequality for all $A \subset X$, and $C \subset X$ is closed. Then $\overline{f[C]} \subset f[\overline{C}] = f[C] \subset \overline{f[C]}$, showing that $f[C]$ is closed.

So we have equality for all $A$ exactly when $f$ is closed and continuous, so we just have to find non-closed continuous maps. And this is what the other answers provide...

Answer 2

Let $A=\mathbb{Z}_+\subset \mathbb{R}_+$ with the standard topology, and let $f(x) = \dfrac{1}{x}$.

The closure of $A$ is $A$, so the image of $f$ does not change. The closure of $f(A)$ is $\{0\} \cup f(A)$.

Answer 3

Take $X$ to be the interval $[1,\infty)$ and $Y$ to be $[0,1]$. Now take the subset A to be X.The function here is $1/x$. So $f(A)$ is $(0,1]$. A is already closed. But $f(A)$ closure is $[0,1]$ . Here the main idea is to create function such that the points far in the domain have images very close in the co domain.So an obvious choice of such function is $1/x$